Trigonometry Examples

$\frac{x^{3} - 27}{x - 3} = x^{2} + 3 x + 9$

Step 1

Factor each term.

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Step 1.1

Rewrite $27$ as $3^{3}$ .

$\frac{x^{3} - 3^{3}}{x - 3} = x^{2} + 3 x + 9$

Step 1.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 3$ .

$\frac{(x - 3) (x^{2} + x \cdot 3 + 3^{2})}{x - 3} = x^{2} + 3 x + 9$

Step 1.3

Simplify.

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Step 1.3.1

Move $3$ to the left of $x$ .

$\frac{(x - 3) (x^{2} + 3 x + 3^{2})}{x - 3} = x^{2} + 3 x + 9$

Step 1.3.2

Raise $3$ to the power of $2$ .

$\frac{(x - 3) (x^{2} + 3 x + 9)}{x - 3} = x^{2} + 3 x + 9$

Step 1.4

Reduce the expression by cancelling the common factors.

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Step 1.4.1

Reduce the expression $\frac{(x - 3) (x^{2} + 3 x + 9)}{x - 3}$ by cancelling the common factors.

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Step 1.4.1.1

Cancel the common factor.

$\frac{(x - 3) (x^{2} + 3 x + 9)}{x - 3} = x^{2} + 3 x + 9$

Step 1.4.1.2

Rewrite the expression.

$\frac{x^{2} + 3 x + 9}{1} = x^{2} + 3 x + 9$

Step 1.4.2

Divide $x^{2} + 3 x + 9$ by $1$ .

$x^{2} + 3 x + 9 = x^{2} + 3 x + 9$

Step 2

Solve the equation.

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Step 2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.1.1

Subtract $x^{2}$ from both sides of the equation.

$x^{2} + 3 x + 9 - x^{2} = 3 x + 9$

Step 2.1.2

Subtract $3 x$ from both sides of the equation.

$x^{2} + 3 x + 9 - x^{2} - 3 x = 9$

Step 2.1.3

Combine the opposite terms in $x^{2} + 3 x + 9 - x^{2} - 3 x$ .

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Step 2.1.3.1

Subtract $x^{2}$ from $x^{2}$ .

$3 x + 9 + 0 - 3 x = 9$

Step 2.1.3.2

Add $3 x + 9$ and $0$ .

$3 x + 9 - 3 x = 9$

Step 2.1.3.3

Subtract $3 x$ from $3 x$ .

$0 + 9 = 9$

Step 2.1.3.4

Add $0$ and $9$ .

$9 = 9$

Step 2.2

Since $9 = 9$ , the equation will always be true for any value of $x$ .

All real numbers

Step 3

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$